Problem Set 5   -   Solutions

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Find the work done when an object is moved along the curve of intersection   C   of the circular paraboloid   z = x ² + y ²   and the plane   2x + y = 2   from (1, 0, 1) to (0, 2, 4) by a force .

2x + y = 2       Þ     y = 2(1 - x)
Let   x = t   then   y = 2(1 - t)     (from the equation of the plane)
and   z = t² + 4(1 - t) ²     (from the equation of the paraboloid)

The value of   t   at   (1, 0, 1)   is   1.
The value of   t   at   (0, 2, 4)   is   0.

Therefore, on the curve   C,




Note the order of the limits of integration.
The path is from the point where t = 1 to the point where t = 0.
Therefore

W = 3 Nm

[Note:

Therefore   F   is not conservative and the value of     depends on the path taken.]

Find the work done in travelling in     once anti-clockwise around the unit circle   C, centered on the origin, in the presence of the force   , using both of the following methods:

1. by direct evaluation of the line integral and

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   A parameterization for the unit circle   C   is
   
   
   
   

   

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2. by using Green’s theorem and evaluating .

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   F   has no singularities anywhere on   , so, by Green’s theorem,

   

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3. Is the vector field   F   conservative?

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       Therefore

   NO,     F is not conservative.

In class we saw that the line integral     has the value   2p, where     and the path   C   is one anti-clockwise circuit of the unit circle centered on the origin.

1. Show that the value of the line integral remains   2p   when the vector field is replaced by
   

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   A parameterization for the unit circle   C   is
   
   
   
   
   

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2. Show that the value of       (where   D   is the circular area enclosed by   C) is not 2p   unless   n < 2.

   Hint:   transform the area integral into plane polar coordinates, with
   x = r cos q,       y = r sin q       and     dA = dx dy = r dr dq.

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   By symmetry,  
   
   By symmetry,  
   
   
   
   
   
   

   Therefore the integral ¹ 2p   unless n < 2.

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3. Parts (a) and (b) clearly show that Green’s theorem does not apply when n > 2.
   Why should Green’s theorem not be used when   n > 0 ?

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   Green’s theorem should not be used if   C   or   D   contain any singularities.
   There is a singularity at the origin (r = 0) if   n > 0.
   There are no singularities anywhere in otherwise.
   Therefore do not use Green’s theorem for   n > 0.

Find a potential function for the vector field  
Use this potential function to evaluate the line integral for any piecewise smooth simple curve   C   from the point (1, 1, -1) to the point (2, 0, 5).

Match this with  
One of several valid solutions is:

where   g(y, z)   is an arbitrary function of   y   and   z.
This expression for   f   leads to  

Þ     g   is a function of   z   alone,     g(z).


Þ     g = A, an arbitrary constant.     Therefore the potential function is

f (x, y, z)   =   x2y cos z   +   A

[The fact that we were able to find a potential function     Þ     F   is conservative.]

C   is any path from   (1, 1, -1)   to   (2, 0, 5).

= (2²×0×cos 5 + A)   -   (1²×1×cos(-1) + A)       Therefore

In the evaluation of an area integral     or a volume integral   ,   the Cartesian differentials are related to the differentials of a different parameterization (u, v, w) of the surface or volume by the Jacobian determinant:

and   ,
where the Jacobian is
.

Evaluate the Jacobian in order to determine the conversion formula from Cartesian differentials   dx dy   (or   dx dy dz) to

1. Plane polar area differentials   dr dq

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   x = r cos q ,       y = r sin q
   

   

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2. Cylindrical polar volume differentials   dr   df   dz

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   x = r cos f,     y = r sin f,     z = z
   
   

   

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3. Spherical polar volume differentials   dr dq   df

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   x = r sin q cos f,     y = r sin q sin f,     z = r cos q
   
   =   | cos q (r² sin q cos q (cos²f + sin²f) ) + r sin q (r sin²q (cos²f + sin²f) ) + 0 |
   =   | r² sin q (cos²q + sin²q) |   =   r² sin q
   [Note that   sin q ³ 0   because   0 £ q £ p]

   

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4. Parabolic area differentials   du dv, where   x = u v,   y = v ² - u ².

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   x = uv,     y = v ² - u ²
   

   

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5. In part (d), sketch on the same x-y plane any three members of each of the two families of coordinate curves   u = constant   and   v = constant.

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   Choosing (arbitrarily) the values   1/2, 1 and 2   for each of   u   and   v:
   
   
   [Note that   u = 0   is the positive y-axis and   v = 0   is the negative y-axis.
   u = c   is the same curve as   u = -c.
   v = c   is the same curve as   v = -c.
   However,   u   and   v   must be of opposite sign wherever   x < 0
   and   u   and   v   must be the same sign wherever   x > 0.]